[tex]F=ma[/tex]
Prin derivare succesiva (sau prin formule daca le stii), se obtine viteza si apoi acceleratia:
[tex]v=50\sqrt{3}\left(\cos 10t+\frac{1}{\sqrt{3}}\sin 10t\right)\\ \\ \\ a=500\sqrt{3}\left(-\sin 10t+\frac{1}{\sqrt{3}}\cos 10t\right)[/tex]
Acum, forta va fi maxima atunci cand acceleratia e maxima. Iat maximul unei functii il aflam prin derivare, ca la matematica:
[tex]a'=5000\sqrt{3}\left(-\cos 10t-\frac{1}{\sqrt{3}}\sin 10t\right)\\ \\ \Rightarrow \cos 10t=-\frac{1}{\sqrt{3}}\sin 10t\\ \\ \tan 10t=-\sqrt{3}\\ \\10t=-\frac{\pi}{3}.[/tex]
Revenind gasim acceleratia maxima inlocuind 10t cu valoarea gasita si calculand:
[tex]a_{max}=500\sqrt{3}\left[-\left(-\frac{\sqrt{3}}{2}\right)+\frac{1}{\sqrt{3}}\frac{1}{2}\right]=\\ \\ \\ =250\cdot 3+250=1000.[/tex]
Iar apoi forta maxima e:
[tex]F_{max}=ma_{max}=0,020\cdot 1000=20.[/tex]
